If a man of ,mass M jumps to the ground from a height h and his centre of mass moves a distance x in the time taken by him to 'hit' the ground the average force acting on him (assuming his retardation to be constant during his impact with the ground) is :

Options

A. `(Mgh)/(x)`
B. `(Mgx)/(h)`
C. `Mg((h)/(x))^(2)`
D. `Mg((x)/(h))^(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the scenario
A man of mass \( M \) jumps from a height \( h \) to the ground. We need to find the average force acting on him during his impact with the ground, given that his center of mass moves a distance \( x \) in the time taken to hit the ground.

### Step 2: Use the Work-Energy Theorem
According to the Work-Energy Theorem, the work done by all forces is equal to the change in kinetic energy.

### Step 3: Calculate the potential energy at the height \( h \)
The potential energy (PE) when the man is at height \( h \) is given by:
\[ PE = Mgh \]

### Step 4: Determine the kinetic energy just before hitting the ground
When the man reaches the ground, all the potential energy will be converted into kinetic energy (KE). Therefore, the kinetic energy just before hitting the ground is:
\[ KE = Mgh \]

### Step 5: Relate work done to force and distance
The work done by the average force \( F \) during the impact can be expressed as:
\[ \text{Work} = F \cdot x \]
where \( x \) is the distance moved by the center of mass during the impact.

### Step 6: Set up the equation using the Work-Energy Theorem
According to the Work-Energy Theorem:
\[ F \cdot x = \Delta KE \]
Since the man comes to rest after hitting the ground, the change in kinetic energy is equal to the kinetic energy just before impact:
\[ F \cdot x = Mgh \]

### Step 7: Solve for the average force \( F \)
Rearranging the equation to solve for \( F \):
\[ F = \frac{Mgh}{x} \]

### Conclusion
Thus, the average force acting on the man during his impact with the ground is:
\[ F = \frac{Mgh}{x} \]